Answer :
Certainly! Let's go through each part of the question step by step.
### Step 1: Calculate the Percentage Nitrogen in [tex]\( \text{Mg(NO}_3\text{)}_2 \)[/tex]
To determine the percentage of nitrogen in magnesium nitrate, we need to consider the molar masses involved.
- The molar mass of magnesium nitrate ([tex]\(\text{Mg(NO}_3\text{)}_2\)[/tex]) is approximately 148.32 g/mol.
- Each nitrogen atom has a molar mass of approximately 14.01 g/mol.
- There are 2 nitrogen atoms in magnesium nitrate.
The mass of nitrogen in one mole of [tex]\(\text{Mg(NO}_3\text{)}_2\)[/tex] is [tex]\(2 \times 14.01 = 28.02\)[/tex] g.
To find the percentage of nitrogen, we use the formula:
[tex]\[ \text{Percentage of Nitrogen} = \left(\frac{\text{Mass of Nitrogen in } \text{Mg(NO}_3\text{)}_2}{\text{Molar Mass of } \text{Mg(NO}_3\text{)}_2}\right) \times 100 \][/tex]
[tex]\[ \text{Percentage of Nitrogen} = \left(\frac{28.02}{148.32}\right) \times 100 \approx 18.89\% \][/tex]
So, the percentage of nitrogen in [tex]\(\text{Mg(NO}_3\text{)}_2\)[/tex] is approximately 18.89%.
### Step 2: Calculate the Concentration of the Nitric Acid that Reacts
We are given that the nitric acid has a concentration of 128 mol/L.
### Step 3: Calculate the Mass of Magnesium that Reacts
From the given balanced reaction:
[tex]\[ \text{Mg} + 2\text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2 \][/tex]
We know 1 mole of magnesium reacts with 2 moles of nitric acid. Given the concentration of nitric acid is 128 mol/L, we can calculate the moles of magnesium that react:
[tex]\[ \text{Moles of Mg} = \frac{128}{2} = 64 \text{ moles} \][/tex]
The molar mass of magnesium ([tex]\(\text{Mg}\)[/tex]) is approximately 24.305 g/mol.
The mass of magnesium reacting can be found by:
[tex]\[ \text{Mass of Mg} = 64 \text{ moles} \times 24.305 \text{ g/mol} = 1555.52 \text{ g} \][/tex]
So, the mass of magnesium that reacts is approximately 1555.52 grams.
### Step 4: Calculate the Volume of Hydrogen Gas Produced at 570K
Given the volume of hydrogen gas produced is 570 dm³ at 570K. To find the volume at standard temperature (273.15K), we employ the ideal gas law properties, specifically using the ratio of volumes to temperatures (as pressure and volume are proportional to temperature):
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Where:
- [tex]\( V_1 = 570 \text{ dm}^3 \)[/tex]
- [tex]\( T_1 = 570 \text{ K} \)[/tex]
- [tex]\( T_2 = 273.15 \text{ K} \)[/tex]
Thus,
[tex]\[ V_2 = V_1 \times \frac{T_2}{T_1} = 570 \text{ dm}^3 \times \frac{273.15 \text{ K}}{570 \text{ K}} \approx 273.15 \text{ dm}^3 \][/tex]
So, the volume of hydrogen gas produced at standard temperature and pressure (STP) is approximately 273.15 dm³.
### Step 5: Calculate the Number of Hydrogen Atoms Present in Nitric Acid
Each molecule of nitric acid ([tex]\(\text{HNO}_3\)[/tex]) contains 1 hydrogen atom.
Given:
- The concentration of nitric acid is 128 mol/L.
- Using Avogadro's number ([tex]\(6.022 \times 10^{23}\)[/tex] atoms per mole).
The total number of hydrogen atoms is:
[tex]\[ \text{Number of hydrogen atoms} = 128 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \times 2 \][/tex]
[tex]\[ \text{Number of hydrogen atoms} \approx 1.541632 \times 10^{26} \][/tex]
Thus, the number of hydrogen atoms present in the nitric acid is approximately [tex]\(1.541632 \times 10^{26}\)[/tex].
In summary:
1. The percentage of nitrogen in [tex]\( \text{Mg(NO}_3\text{)}_2 \)[/tex] is approximately 18.89%.
2. The concentration of the nitric acid that reacts is 128 mol/L.
3. The mass of magnesium that reacts is approximately 1555.52 grams.
4. The volume of hydrogen gas produced at 570K is 273.15 dm³ at standard temperature.
5. The number of hydrogen atoms present in nitric acid is approximately [tex]\(1.541632 \times 10^{26}\)[/tex].
### Step 1: Calculate the Percentage Nitrogen in [tex]\( \text{Mg(NO}_3\text{)}_2 \)[/tex]
To determine the percentage of nitrogen in magnesium nitrate, we need to consider the molar masses involved.
- The molar mass of magnesium nitrate ([tex]\(\text{Mg(NO}_3\text{)}_2\)[/tex]) is approximately 148.32 g/mol.
- Each nitrogen atom has a molar mass of approximately 14.01 g/mol.
- There are 2 nitrogen atoms in magnesium nitrate.
The mass of nitrogen in one mole of [tex]\(\text{Mg(NO}_3\text{)}_2\)[/tex] is [tex]\(2 \times 14.01 = 28.02\)[/tex] g.
To find the percentage of nitrogen, we use the formula:
[tex]\[ \text{Percentage of Nitrogen} = \left(\frac{\text{Mass of Nitrogen in } \text{Mg(NO}_3\text{)}_2}{\text{Molar Mass of } \text{Mg(NO}_3\text{)}_2}\right) \times 100 \][/tex]
[tex]\[ \text{Percentage of Nitrogen} = \left(\frac{28.02}{148.32}\right) \times 100 \approx 18.89\% \][/tex]
So, the percentage of nitrogen in [tex]\(\text{Mg(NO}_3\text{)}_2\)[/tex] is approximately 18.89%.
### Step 2: Calculate the Concentration of the Nitric Acid that Reacts
We are given that the nitric acid has a concentration of 128 mol/L.
### Step 3: Calculate the Mass of Magnesium that Reacts
From the given balanced reaction:
[tex]\[ \text{Mg} + 2\text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2 \][/tex]
We know 1 mole of magnesium reacts with 2 moles of nitric acid. Given the concentration of nitric acid is 128 mol/L, we can calculate the moles of magnesium that react:
[tex]\[ \text{Moles of Mg} = \frac{128}{2} = 64 \text{ moles} \][/tex]
The molar mass of magnesium ([tex]\(\text{Mg}\)[/tex]) is approximately 24.305 g/mol.
The mass of magnesium reacting can be found by:
[tex]\[ \text{Mass of Mg} = 64 \text{ moles} \times 24.305 \text{ g/mol} = 1555.52 \text{ g} \][/tex]
So, the mass of magnesium that reacts is approximately 1555.52 grams.
### Step 4: Calculate the Volume of Hydrogen Gas Produced at 570K
Given the volume of hydrogen gas produced is 570 dm³ at 570K. To find the volume at standard temperature (273.15K), we employ the ideal gas law properties, specifically using the ratio of volumes to temperatures (as pressure and volume are proportional to temperature):
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Where:
- [tex]\( V_1 = 570 \text{ dm}^3 \)[/tex]
- [tex]\( T_1 = 570 \text{ K} \)[/tex]
- [tex]\( T_2 = 273.15 \text{ K} \)[/tex]
Thus,
[tex]\[ V_2 = V_1 \times \frac{T_2}{T_1} = 570 \text{ dm}^3 \times \frac{273.15 \text{ K}}{570 \text{ K}} \approx 273.15 \text{ dm}^3 \][/tex]
So, the volume of hydrogen gas produced at standard temperature and pressure (STP) is approximately 273.15 dm³.
### Step 5: Calculate the Number of Hydrogen Atoms Present in Nitric Acid
Each molecule of nitric acid ([tex]\(\text{HNO}_3\)[/tex]) contains 1 hydrogen atom.
Given:
- The concentration of nitric acid is 128 mol/L.
- Using Avogadro's number ([tex]\(6.022 \times 10^{23}\)[/tex] atoms per mole).
The total number of hydrogen atoms is:
[tex]\[ \text{Number of hydrogen atoms} = 128 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \times 2 \][/tex]
[tex]\[ \text{Number of hydrogen atoms} \approx 1.541632 \times 10^{26} \][/tex]
Thus, the number of hydrogen atoms present in the nitric acid is approximately [tex]\(1.541632 \times 10^{26}\)[/tex].
In summary:
1. The percentage of nitrogen in [tex]\( \text{Mg(NO}_3\text{)}_2 \)[/tex] is approximately 18.89%.
2. The concentration of the nitric acid that reacts is 128 mol/L.
3. The mass of magnesium that reacts is approximately 1555.52 grams.
4. The volume of hydrogen gas produced at 570K is 273.15 dm³ at standard temperature.
5. The number of hydrogen atoms present in nitric acid is approximately [tex]\(1.541632 \times 10^{26}\)[/tex].
